Question

In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053nm around a stationary proton.

How many revolutions per second does the electron make (Answer in Hz)? Hint: What must be true for a force that causes circular motion?

I'm very confused about how to even begin, and there's no example in my Physics book to refer to! Please help if you can! Thanks!

Answer 1

In case you exrpess charge of proton and charge of electron in SI metric system, you should write the formula for the electric attraction force in SI system too
F = -kq1q2/r^2
From the other side, F = ma = mv^2/r
And v is the velocity v = 2rPi/T, where T is the Time of one revolution. The amount of revolutions per second n equals 1/T, so
F = mv^2/r = m(2xPixrxn)^2/r = 4Pi^2mrn^2 from one side, and
F = -kq1q2/r^2 from other side, so
4Pi^2mrn^2 = -kq1q2/r^2
and n = sq.root{-kq1q2/(4Pi^2r^3m)} =
= q1/(2Pir)sq.root{(c^2/10^7)/rm} = (every values we shall express in SI system) = q1c/(2Pir)sq.root{1/10^7rm} = 1.602x10^-19x3.0x10^8/(6.28x0.53x10^-10)… =
= (4.806x10^-11/3.328x10^-10)x0.4551x10^17… 0.658x10^16 revolutions per second.

To check the result, let us find the energy of ionization of hydrogen atom and find the frequency of ionizing photon E=hn(l). Hydrogen ionization energy is (using the enciclopedia data)
E=13.595(electronvolts)=13.595x1.602x1…
In case we divide this energy by h, we shall obtain
n(l) = E/h = (2.178/6.626)10^-18x10^34 = 0.3287x10^16 periods per second
We see that this frequency is just one half of the previous. From general consideration they should be equal.
I have checked the calculations, they are correct. Possibly, there should be some quantum consideration to obtain the frequency. Yet, you can use the first obtained frequency as the result for the question of the problem under classic rotation laws.